Question

The gravitational force of attraction between a uniform sphere of mass $M$ and a uniform rod of length / and mass $m$ oriented as shown is

• A. $\frac{GMm}{r(r+l)}$
• B. $\frac{GM}{r^2}$
• C. $Mm{r}^2+L$
• D. $\left(r^2+L\right)mM$

Answer 1

Answer: (A)

The gravitational field of $M$ at $x=\frac{GM}{x^2}$
$dF=\frac{GM}{x^2}dm$
where $dm=\frac{m}{l}dx$
$\Rightarrow dF=\frac{GMmdx}{l\cdot x^2}$
$F=\frac{GMm}{l}{\int }_r^{r+l}\frac{1}{x^2}dx=\frac{GMm}{l}{\left[-\frac{1}{x}\right]}_r^{r+l}=\frac{GMm}{l}\left[\frac{1}{r}-\frac{1}{r+l}\right]$
$F=\frac{GMml}{l\cdot r(r+l)}=\frac{GM\cdot m}{r(r+l)}$